AP's 204th book of Science// 7-Circle-theorem in 2D becomes 9-Band-theorem in 3D. Pi in 2D is 22/7, and equiangular number is 19/7, because of 7 circle theorem. There are in the Plutonium Atom Totality, 22 subshells inside of 7 shells, only 19
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Archimedes Plutonium<plutonium.archimedes@gmail.com>
Aug 30, 2022, 5:14:25 PM
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AP's 204th book of Science// 7-Circle-theorem in 2D becomes 9-Band-theorem in 3D.

Pi in 2D is 22/7, and equiangular number is 19/7, because of 7 circle theorem. There are in the Plutonium Atom Totality, 22 subshells inside of 7 shells, only 19 subshells occupied at any given moment of time.

In 3rd dimension electricity and magnetism becomes band-like structures form toruses. The equivalent of the 7-Circles-theorem in 2nd dimension is the 9-Band-theorem in 3rd dimension.

In this book, my aim is to prove 3rd dimension is the last and final dimension. And that a 4th dimension is merely a summation of two 2nd dimensions. The band-like structure is this summation.

In Algebra we have the limitations of speed of light as no faster than the speed of light and at a constant speed. This of course is Special Relativity.

But now we connect Special Relativity to geometry. We take it further into math where it was confined to algebra, and take Special Relativity into geometry.

When we apply Special Relativity to geometry we end up with Band-like structure and the 9-Band-theorem.

In the 7-Circle-theorem, all given circles can be broken down into 6 interior circles, unique to the 7th all encompassing circle. And we can move those 6 interior circles around at a speed obeying Special Relativity.

In the 9-Band-theorem, with 9 circle-bands in 3rd dimension. If we make the bands be circles there are 14 circles in all in 3rd dimension. In 2D, we had only circles, in 3D we double the circles to form a torus of circle bands.

Every arbitrary given circle is broken down into 6 interior unique circles.

Every arbitrary circle-band torus is broken down into 14 unique circles.

Special Relativity as we know it, is a 2D phenomenom as we stick to speed in 2D.

Special Relativity in 3D changes for the geometry is different and in 3D, Special Relativity involves the torus, a torus of 9-Circle-Bands.

AP, King of Science, especially physics
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Archimedes Plutonium<plutonium.archimedes@gmail.com>
Aug 30, 2022, 5:41:52 PM (8 hours ago)
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What Special Relativity does to 2nd dimension is place limitations on speed..

What Special Relativity does to 3rd dimension is place limitations on dimension itself, in that 4th dimension is a composition of two 2nd dimensions.

And this is what magnetic field and electric field are. Magnetic field is a 2nd dimension parameter. Electric field or angular momentum is a 2nd dimension parameter. When you multiply current by magnetic field by electric field you get Voltage, a 3D parameter.

AP
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Archimedes Plutonium<plutonium.archimedes@gmail.com>
Aug 30, 2022, 8:17:56 PM (5 hours ago)
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Just starting off and I already made a big mistake. But it is best to make mistakes at the beginning, so they are easily corrected straight away.

The mistake is not all entirely my fault, but a misconception by Old Math in naming their theorem. They named it the 7-circle theorem when a far better name would have been Hexagon Circle Theorem, on the fact that the 6 interior circles are where the action is. And a circumscribed hexagon interior leads to a fast simple proof 7-Circle.

But if we rename 7 Circle to be the more appropriate name of 6-Circle theorem. Then the 9-Band-theorem I wrote about earlier today makes more sense, more commonsense, because the spacing of 8 band rings to form a torus and then the spacing of 4 identical band rings on the equator inside the 8, gives a total of 12 band rings. And so we have a 12-Band-Ring-Torus theorem, the 3rd dimensional equivalent of the 6-Circle theorem of 2nd dimension. Both are universal for the torus and for the circle, respectively.

Glad I caught this mistake of mine early on.

P.S. for a visual of a torus composed of band rings , see the cover picture of either of these Amazon Kindle published books:

My 148th published book

Plutonium Atom Totality Universe, 9th edition 2021, Atom Totality Series, book 1 Kindle Edition
by Archimedes Plutonium (Author)

Last revision was 30Nov2021. And this is AP's 148th published science book.

Preface:
Physics book that explains what the universe is, and how it works. This is a continuation of the Atomic Theory by Democritus in Ancient Greek times. It adds one more fact to the Atomic Theory picture. That the Universe itself is one gigantic big atom. The picture is that of the Universe, in total, is one big atom that contains more atoms, inside itself. It completes the logic of science that Dr. Feynman wrote-- all things are made up of atoms -- and so, to complete that idea -- all things and the universe itself is an atom.

In this edition of year 2021, AP actually proves the Atom Totality theory, and therefore, a simultaneously _disproof_ of the Big Bang theory. The proof is simple, in that the Sun and stars shine not from fusion, but rather, instead, shine from the Faraday law going on inside each and every proton in the Sun or star, or, in the Universe. The muon is the true electron of atoms and is the bar magnet in Faraday's law while the proton is a 840MeV particle in the geometry shape of 8 ring coil torus that is the closed loop coil in Faraday's Law. NASA scientists have discovered the Sun is a yearly increase in radiation of 0.005% yearly, and is why 25% of all insects have perished in the last decade 2010-2020. Stars and Sun shine from Faraday law, not from fusion and that is how the Universe itself grows. So, we cannot logically have two different mechanisms for the creation of the Universe. We cannot have electricity magnetism of Faraday law and then some silly "explosion of Big Bang" to create and grow the universe.

Also, in the course of providing supporting evidence of the Atom Totality theory, my research had to revise and correct the entire Maxwell Equations, and revise and correct the theory, the quantum electrodynamics theory.

Cover Picture: Again I used 8 rings from plumbing hardware to represent the 8 rings of a proton torus, and visualize each ring as a dot cloud pattern instead of a continuous ring, and the holes in some of those rings helps facilitate that image. Notice the muon ring is inside the proton torus rings, and perpendicular, and situated at the equator, going around and around the proton torus at nearly the speed of light in the Faraday law; producing electricity. Atoms are designed to produce maximum electricity, given their masses.
Length: 329 pages

Product details
• ASIN ‏ : ‎ B08T82M2LP
• Publication date ‏ : ‎ January 16, 2021
• Language ‏ : ‎ English
• File size ‏ : ‎ 872 KB
• Text-to-Speech ‏ : ‎ Enabled
• Screen Reader ‏ : ‎ Supported
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Enabled
• Print length ‏ : ‎ 329 pages
• Lending ‏ : ‎ Enabled
• Best Sellers Rank: #203,710 in Kindle Store (See Top 100 in Kindle Store)
◦ #289 in Physics (Kindle Store)
◦ #1,716 in Physics (Books)

My 135th published book

QED in Atom Totality theory where proton is a 8 ring torus and electron = muon inside proton doing Faraday Law// Atom Totality series, book 6 Kindle Edition
by Archimedes Plutonium (Author) 

Since the real true electron of atoms is the muon and is a one ring bar magnet thrusting through the 8 ring torus of a proton, we need a whole entire new model of the hydrogen atom. Because the Bohr model with the 0.5MeV particle jumping orbitals as the explanation of Spectral Lines is all wrong. In this vacuum of explaining spectral line physics, comes the AP Model which simply states that the hydrogen atom creates Spectral lines because at any one instant of time 4 of the 8 proton rings is "in view" and the electricity coming from those 4 view rings creates spectral line physics.

Cover Picture: Is a imitation of the 8 ring proton torus, with my fingers holding on the proton ring that has the muon ring perpendicular and in the equatorial plane of the proton rings, thrusting through. This muon ring is the same size as the 8 proton rings making 9 x 105MeV = 945MeV of energy. The muon ring has to be perpendicular and lie on the equator of the proton torus. Surrounding the proton-torus would be neutrons as skin or coating cover and act as capacitors in storing the electricity produced by the proton+muon.

Product details
• ASIN ‏ : ‎ B08K47K5BB
• Publication date ‏ : ‎ September 25, 2020
• Language ‏ : ‎ English
• File size ‏ : ‎ 587 KB
• Text-to-Speech ‏ : ‎ Enabled
• Screen Reader ‏ : ‎ Supported
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Print length ‏ : ‎ 25 pages
• Lending ‏ : ‎ Enabled
• Best Sellers Rank: #240,837 in Kindle Store (See Top 100 in Kindle Store)
◦ #8 in 45-Minute Science & Math Short Reads
◦ #35 in General Chemistry & Reference
◦ #264 in General Chemistry

In Wikipedia on 7-Circles theorem says it was not discovered until 1974 by Evelyn, Money-Coutts, and Tyrrell.

Without looking at the proof, I would almost bet that their proof is false because it cannot prove the center of the big circle is captured by the diameters of the other 6 circles. And how could they since pi in Old Math is not ending in zeroes digits.

So here is a case in point, that one can possibly see if 3.14000.... or 3.160000..... can be used to prove this theorem.

It is the pi number 3.14159.... never ending in zero digits, that makes me think this is squaring pi type of proof, and is invalid.

So I am even skeptical that there is a "valid" proof of 7-Circles theorem. Because of the all sizes of those 6 interior circles.

On the other hand, when AP asks for a _Regular Hexagon_ instead of _Irregular Hexagon_ be circumscribed, then a proof can ascertain where exactly the center of the largest circle is.

So I start with a regular hexagon inside a circle, then I build out the circles, or ovals, or even ellipses (short stubby ellipses).

So, well, I almost bet there was never a valid proof of 7-Circles theorem, due to pi cannot handle a division by 6.

However, if you start with the Regular Hexagon and locate its center, then, can 6 smaller circles at tangent to its neighbors. That would be a proof.

And I am looking at a regular hexagon at this moment-- and yes, I can easily see 6 smaller all the same size circles intersecting and also being the diameter center point of the large circle.

So I suspect 7-Circle theorem is a fakery, all because the center cannot be guaranteed. And the reason being-- pi number does not end in zero digits.

AP

Now recently I got screwed up on a geometry proof, because of Computer Graphics on the Goldberg polyhedron. That computers can easily deceive geometers. I had thought there can be a 6th regular polyhedron based on the hexagon.. And when you look at the Goldberg polyhedron, he skips and glides over the fact that hexagons have "no bend" to them to go around on a sphere surface. But a computer graphic can easily deceive the geometer, as I was optically deceived into thinking Goldberg had hexagons tile a sphere surface.

Now the 7-Circle theorem, and let me call it a fakery, is said to date to 1974.

What does the Goldberg polyhedron optical illusions date to??? Wikipedia only mentions the date of first description as 1937.

What I am looking for is when computer graphics entered into geometry mathematics and installed this massive optical illusions.

Can one say that both 7-Circle and Goldberg polyhedron are total trash mathematics, brought about by the insidious computer graphics that so easily brings in optical illusions, and hence Fake theorems.

If so, well every geometry proof since 1937 and beyond needs to be inspected for validity. For the optical illusions that computers can cast upon geometry figures is more harmful than beneficial. Where we end up needing "hands on of geometry figures" for valid proofs.

AP

Alright, my mistake, the three chords do not have to meet at the large circle center. So it appears this theorem could be valid. However, let AP add another Lemma on top of this theorem-- the only center intersection is a set of identical circles, all 6 of them, and the regular-hexagon circumscribed by the larger circle. All other intersections will not meet at the center of the largest circle.

AP

Alright, now that I fully know what the 7-Circle theorem is about, moving along to the 3D version. For the 2D version seems rather unimportant and insignificant.

And of course, I will not write up my flubb-ups.

But it is still of major concern that computer graphics can make optical illusions that destroy a proof, and procure fakeries.

I read the proof but still am not sure how they determined the 3 chords intersect in one point. I think they used a triangle proof of circles to justify the intersection. And that could be rife for computer graphics optical illusion and fake proof. I am not going to dig much deeper into the 7 -Circle-theorem.

Now looking at the Wikipedia entry again what if we had three large identical circles and all three intersect not only the large outer circle but themselves, and in between three small identical circles to round out the 6 inner circles. Could it be that this chord intersection is the large circle center?

So the 7Circle theorem is losing its excitement and luster, and bogged down in questions of validity.

For what I want out of this is more understanding of the torus.

I may call this book off altogether, as the 7 Circle is beginning to disconnect from torus.

AP

Almost impossible for AP to leave a math proof when dug in so deep already. I suspect I found where this 7-Circle theorem is a fraud.
Archimedes Plutonium<plutonium.archimedes@gmail.com>
Aug 31, 2022, 4:34 AM
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I changed my mind and calling this book off. It no longer interests me. At first I thought the 7-Circle-Theorem was unique to 6 interior circles of a larger circle and that they involved their diameters and not chords. I also thought the statement of the theorem said the 3 line segments intersected the largest circle at its center.

So I thought the 6 inner circles were a set unique to every given circle and passed through the center of large circle.

That is not the 7-Circle theorem. And so, it is not interesting. It has no connection to a torus in 3rd dimension.

I am looking for a better understanding of the torus, not some minor issues of 6 circles inside a large circle.

If the 7 Circle theorem had been this-- given any circle, there exists 6 unique smaller circles that intersect the large circle and intersect each neighbor and whose chord diameter intersects the center of the large circle, then we have something highly interesting.

AP

Alright, I am going to continue with this book if it becomes established as a fake theorem. It is almost impossible for me to leave a subject of math after digging in this deep.

Take a look at Wikipedia's page of the Seven Circle Theorem, although a bit deceptive in and of itself for one has to look hard to realize it is a Irregular Hexagon, although it looks almost like a Regular Hexagon. And Wikipedia's diagram is deceptive also in the idea that the chords look like diameters, not chords. So a very deceptive picture by Wikipedia led me to figure out what the theorem was all about without even reading a statement of the theorem, only to find out, it was nothing what I thought it was.

The statement of 7-Circle theorem involves only the ideas of those chords meeting in a single point, often not the center of the largest circle.

I have not pinpointed how mathematicians were beguiled into believing those chords meet into a point. But I am guessing that they used another theorem which is also fake, concerning the 3 circles or 5 circles in a triangle. So the mistake maybe, I am not sure, the mistake maybe using a fake proof of another theorem to make the 7-Circle theorem. Sort of like compounding a fake proof to be more fake proofs.

Now in the Dandeloin fake proof of a two spheres balanced on a ellipse inside a right circular cone, the fake proof is easily seen where the fakery is.. That Dandeloin ASSUMED the slant cut was a ellipse with 2 foci. Yet in truth the slant cut of a cone is a Oval, with no foci. So Dandeloin fake proof of math is where you actually end up assuming something is true from the start when it is not. And the proof is just window dressing.

In the case of 7-Circle fakery, I doubt it was a fakery from a Assumption. Other than assuming the use of a previous theorem is true or fakery.

So, here is AP's take on this 7-Circle theorem that indicates, strongly indicates it is fakery.

Indications 7-Circle is a Fake Proof, a Fake theorem.

(1) It is proven, well known proof that 3 points in a plane makes for a Unique circle. That is well proven.

(2) It is seen that in the 7-Circle, that the interior of the largest of the 7 circles harbors 6 smaller circles of various sizes but all 6 intersect the largest circle at one point and all 6 intersects its two neighboring circles at one point each.

(3) Thus the inner 6 circles forms a hexagon. And most examples of 7-Circle theorem are Irregular Hexagons. (AP wrote a proof yesterday that a Regular Hexagon obeys the 7-Circle theorem and the point of intersection is the center of the largest circle.)

(4) Since the interior 6 circles forms a hexagon, we can take 3 points of one side of the hexagon and compare with 3 points of the other side of the hexagon.

(5) If the 3 points on one side of the irregular hexagon produces the same circle as the 3 points of the other side of the irregular hexagon, then there is a Chance that the 3 chords meet in a point. But if the 3 points of one side do not yield the same unique circle as the other side, then the 3 chords do not intersect at one point.

I suspect the only hexagon in which one set of 3 sides can produce the same unique circle is the Regular Hexagon.

And here I need to look up whether there is a Old Math proof that in a Irregular Hexagon on whether the diagonals meet at a single point of intersection. This begs the question of how Irregular must a hexagon be, before the 3 diagonals miss each other in a one-point-intersection.

Mind you, at this point, I have not proven 7-Circle Theorem is a fakery, for the constraints of the circles make be enough constraints to make one set of 3 points converge on the same unique large circle as the other set of 3 points of the Irregular Hexagon.

If the 7 Circle theorem turns out to be true, then what it does, is place a boundary of where Irregular Hexagons are still having some form of Symmetry. But if the 7 Circle theorem is a fakery, then the only symmetrical hexagon is the Regular Hexagon.

AP, King of Science, especially physics

Not looking good for the 7-Circle theorem for in math stack exchange under the title "Diagonals dividing an hexagon into regions of the same area"

Shows and Irregular Hexagon and clearly shows the 3 diagonals as missing one another in a 1 point intersection. Now that diagram is a very or should I say highly Irregular hexagon.

So, we have two possibilities here. If the constraints of the 6 circles inside the 7th large circle are enough constraints to make the diagonals of a "lightly irregular hexagon" to meet in a 1 point intersection.

Or the other possibility-- the only the Regular Hexagon can meet in a 1 point intersection.

So what the 7 Circle Theorem is leading up into, is the question of whether the circle constraints allow for some Symmetry to remain to deliver a 1 point intersection.

Or, what AP feels is going to happen-- that only the Regular Hexagon retains the symmetry that a 1 point intersection occurs. And that only the Regular Hexagon has the symmetry for 3 points of the 6 and the other 3 points of the 6 deliver the same Unique Large circle of the 7th circle in 7-Circle theorem.

I have a tough time in my mind to think that Slightly Irregular Hexagons, that broken into two sets of 3 points, delivers the Same Unique large circle of the 7th circle.

And if AP is correct on this issue, that 7 Circle Theorem is a FAKERY, much the same as Dandeloin is a fakery of geometry and that Goldberg polyhedron are fakery. Then we have a much larger problem on our hands in that Computer Graphics of modern day science is more deceptive and beguiling than it ever is any help in mathematics and science. That we have to throw out all computer aided geometry proofs.

AP

Now I am thinking through a square and rectangle can have all 4 points enclosed inside a circle.

I know a Regular Pentagon and a Regular Hexagon enclosed inside a large circle.

But where does a Irregular Pentagon stop being inside a single large circle?

Where does a Irregular Hexagon stop being inside one large circle as in the 7-Circle Theorem?

Let me go back to 4 sided figures. A square and rectangle can always form a unique large circle. But now, how about quadrilaterals, or parallelograms or trapezoids? Where do these figures stop being all enclosed in one large circle, if any?

AP

On Wednesday, August 31, 2022 at 5:30:37 PM UTC-5, Archimedes Plutonium wrote:
> Now I am thinking through a square and rectangle can have all 4 points enclosed inside a circle.
>
> I know a Regular Pentagon and a Regular Hexagon enclosed inside a large circle.
>
> But where does a Irregular Pentagon stop being inside a single large circle?
>
> Where does a Irregular Hexagon stop being inside one large circle as in the 7-Circle Theorem?
>
> Let me go back to 4 sided figures. A square and rectangle can always form a unique large circle. But now, how about quadrilaterals, or parallelograms or trapezoids? Where do these figures stop being all enclosed in one large circle, if any?
>

So this is a good exercise in general to attack the 7 Circle theorem for validity.

Let us take a irregular square, just as we would take an irregular hexagon for 7 Circle theorem.

Say our Square is a quadrilateral of side 1, 2, 3 and 4 lengths. Can that quadrilateral be enclosed in a unique circle? We know from a theorem, a true theorem that every arbitrary three points is a unique circle. But how about 4 points of lengths 1,2,3,4.

So here I am getting into the theory of geometry of when a quadrilateral ceases to be able to be circumscribed by a circle. This is the heart of the 7 Circle theorem if that theorem is true. For if that theorem is true, all irregular hexagons that obey the rules of 7 Circle theorem are enclosed in a unique large circle, -- the 7th circle.

For quadrilaterals, we know squares form a unique enclosing circle, so do rectangles, so do trapezoids. What about parallelograms, they are 4 sided but do they allow a unique large circle to circumscribe them?

This is the essence of the 7 Circle theorem-- we know a Regular hexagon allows a large unique circle to circumscribe, but we do not know to what extent Irregular hexagons allow for a large circle circumscribe them.

AP

And this is a highly worthwhile exercise of logic and math and geometry in particular. We have a proof that 3 arbitrary points in the plane form a unique circle. So what type of quadrilateral is required to not be able to enclose that quadrilateral inside a circle? We know a square and rectangle and trapezoid form unique circles, and all three of those figures have some parallel symmetry. But where in quadrilaterals do we start __ Not able to form a circumscribing circle around all 4 vertex points of quadrilaterals? Do all parallelograms have unique circles for they have some parallel symmetry?

AP

It has been known for a long long time that 3 arbitrary points in the plane determine a unique circle. Hence, every triangle of its vertices determine a unique circle.

What has not been known is to what extent the 4 vertices of 4 sided polygons determine a unique circle. And what quadrilaterals make it impossible to determine a circle.

This question then bears down on hexagons, and the 7-Circle theorem. If the 7-Circle theorem is true, which is still questionable of its validity, would say that many Irregular Hexagons determine a unique circle-- those hexagons that obey the rules of 7-Circle theorem. But the proof of 7-Circles theorem, the alleged proof uses other theorems that are also questionable validity.

So to get a idea, a estimate if Irregular Hexagons can ever determine a unique circle, we look to quadrilaterals, and see if Irregular quadrilaterals can determine a unique circle.

Tonight I drew a irregular quadrilateral with sides 1,2,3,4 lengths, and I am finding this 4 sided figure of 4 vertices even constructing a circle using those 4 vertices.

But another question pops to my mind. Are there Impossible Quadrilaterals, given arbitrary lengths, and just immediately I can answer that-- yes-- a resounding yes. For a quadrilateral of sides length 1000, 1,2,3 is impossible to exist. But now, is there some rule for when arbitrary side lengths are given, is there some rule which tells us immediately, that those lengths cannot create a quadrilateral. Such a question is likely relevant and linked to the question of when does a quadrilateral impossible to form a circle? Looking at my quadrilateral of 1,2,3,4 lengths of sides, they determine angles of the 4 interior angles. And if those 4 interior angles are _not progressive_ then the ability of the 4 vertices is impossible to create a circle from the 4 vertices.

AP, King of Science, especially physics

Archimedes Plutonium<plutonium.archimedes@gmail.com>
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Alright, so every triangle with its vertices determines a circle.

Every triangle is determined by three distance lengths provided A+B is greater than C, where A,B,C are the lengths of three sides.

Now can this rule also say which lengths are impossible to form a quadrilateral? That A+B+C is greater than D, where A,B,C,D are the distance lengths of quadrilateral four sides.

So in general rules, a polygon exists only if all sides, except one side, when added up is greater than largest side.

Now the number of sides of the polygon and the length of those sides determine the interior angles. And those interior angles have to be progressively larger to create a circle that encompasses all of the vertices.

So all triangles create a unique circle.

Some quadrilaterals create a unique circle. What is the characteristic of quadrilaterals that cannot create a unique circle from their vertices?

AP
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Alright, I have found a easy class of quadrilaterals that are anti-circle building. They resist being circumscribed by a circle of their 4 vertices. A quick picture of one of these 4-gons tells the whole story.

/|
| |
| |
_

You see the antisymmetry of /| with the other end being flat _ Forces the circle creation to be unable to negotiate the /| end with the flat end _.

Now I need to see if I can apply that to irregular hexagons to test the 7-Circle theorem for validity.

AP

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Alright, from preliminaries, looking at the diagram in Wikipedia of their 7-Circle theorem of an irregular hexagon, it appears the 7-Circle-theorem is valid in the sense of encompassing one large circle, the 7th. That, like rectangles or trapezoids or parallelograms in 4 gons as irregular from the square, they still determine a unique large circle. However in the 4-gons with one end pointed and the other end flat, as in my sketch, a circle is impossible to create. So in the 7-Circle theorem, irregular hexagons do create a circle. But we still need to verify the 3 chord intersection as a single point intersection. We are not out of the woods yet on the 7-Circle theorem.

Using the mechanism that opposite sides have to be parallel, satisfies the condition of determining a circumscribing unique circle. But again, it does not necessarily mean the intersection is a single point of the 3 chords, 3 diagonals of irregular hexagon.

It may turn out that some irregular hexagons come close to a single point intersection but miss it narrowly. And Computer Graphics offers a optical illusion that they intersect at one point, when actually they just come close..

Alright, I am closing in on the 7-Circle theorem and it does not look good in terms of Validity. At this moment in time it looks to be Invalid-- mostly a piece of junk.

The last feature I am analyzing is whether they intersect in a single point and that is not looking good, not good at all.

Yesterday I was able to find out what can wreck a circle enclosure. For the PRIMARY Question of 7-Circle is -- can an Irregular Hexagon create a enclosed large circle. That much is true. For a rectangle is an irregular square and it easily creates a enclosed circle about the rectangle.

And the way to make a 4-gon fail to determine a enclosed circle is to examine a rectangle whose one end is pointed and other end is flat.

/|
| |
| |
_

No circle can enclose that.

But now, how about a parallelogram of two pointed ends. There is where I went to bed last night and could not finish answering the question. I went to bed leaving the problem to solve 7-Circle theorem with the question of do the 3 chords actually meet in a single point.

I am always better at logic in the morning than late at night.

And I can safely say, that the 7-Circle-Theorem is a pile of junk.

Usually in mathematics a theorem statement is a Logical IF-> THEN statement..

For example, if you have three arbitrary points in the plane, then, a unique circle is determined by those 3 arbitrary points of the plane.

Now, examine the 7-Circle bs theorem. Examine it logically.

It says-- If you have 6 smaller circles inside a larger circle such that those 6 intersect the larger 7th circle at one point and each inner circle intersects two and only two neighboring inner circles, then, then the chords intersect at a single point.

What the Old Math Statement of 7-Circle theorem was --- was a Logically Incorrect statement of what the theorem should really be. The correct statement of 7-Circle theorem should have been this one.

If you have 6 smaller circles going around in a somewhat circular fashion and every circle connected to its neighbor on right and left by a single point intersection such that those 6 small circles of different sizes or the same size form a hexagon of their outermost point, whether a regular hexagon or an irregular hexagon, then those 6 outermost points determine a unique larger circle.

You see, the mistake of Old Math is that they thought the IF Then of 7-Circle produces a common point intersection of 3 chords. When in reality, the theorem IF THEN produces a 7th circle determined by the 6 smaller circles.

The authors of the original 7-Circle-theorem did not have enough logical reasoning to understand the 3 chord intersection was irrelevant to the theorem itself.

AP

Alright, I have a proof of the 7-Circle-theorem but it is a theorem totally different from the bullshit theorem which is fake announced by Evelyn, Money-Coutts, and Tyrrell of 1974. Their so-called theorem is in tatters, ruins.

The true theorem associated with 7-Circle involves 6 circles and if the conditions are right, then those 6 circles produces or determines a unique 7th circle that encloses the 6. The 1974 bullshit 7-Circle is focused on an irrelevancy of whether chords intersect at one single point. Whether they do or not is irrelevant to producing a 7th new circle unique to the 6 inner smaller circles.

Say you start out with 6 circles all of different sizes. But not too far different in sizes. For example radius 1, 1.1, 1.2, 1.3, 1.4, 1.5.

Then because they are suitable sizes then form a "somewhat circular chain" etching out a large circle. And suppose their size links all 6 of these circles together, of left and right neighbor intersecting at a single point. If those conditions are met, then we find the most exterior point of each of those 6 circles and connect them up into a hexagon. And that hexagon determines a unique large circle, a 7th circle that encloses all 6 smaller circles.

No-one cares about chords intersection, for the theorem is all concerned about when are 6 smaller circles able to Determine a unique 7th larger circle..

Now I was looking to see if long thin parallelograms form a unique circle encompassing all 4 points. And today I take up that project. Apparently thin long parallelograms are the deciding determinant of whether a IRREGULAR 4-gon determines a unique circle.

All squares and rectangles determine a unique circle. But, it is the parallelogram that at some angle of its **needle like ends, wedge like ends** determines if the enclosed rounded figure will be a circle or the next level of roundness, the ellipse or the next level of roundness the oval.

Here is a picture diagram

/|
| |
| |
|/

You see those needle or wedge ends, if the wedge ends are flat they be a rectangle and allow for a circle creation. But if pointed, they still have symmetry. Can they form a circle? Maybe so, maybe not. But the parallelogram is the deciding figure.

Now if we examine a Trapezoid. And place the base on the diameter of a circle, can all trapezoids determine a circle in this fashion? Some can, some cannot, depending on whether the upper base is reasonably close in length to the lower base.

Now we double the Trapezoid and it forms 6 points of intersection with enclosed circle.

Now we ask the question of diagonal intersection. For these are all Irregular (except one) hexagons. These are similar to the 7-Circle theorem of AP. And all the diagonals of all these Trapezoids forming irregular hexagons, all of them meet at the circle center as well as the hexagon center.

Now, what if I make these trapezoids, slanted sides different as in the Old Math 7-Circle theorem. And I mirror reflect the trapezoid in the other semicircle. The question here is, does the 3 diagonals meet at one point intersection?

AP

The AP 7-Circle-Theorem, a correcting of the error filled Old Math 7-Circle-Theorem of 1974.

A statement of the AP 7-Circle-Theorem: The true theorem associated with 7-Circle involves 6 circles and if the conditions are right, then those 6 circles produces or determines a unique 7th circle that encloses the 6. The 1974 bullshit 7-Circle is not focused on the unique determination of the 7th larger circle, and using the intersection as a constraint on Irregular hexagon forming the 7th circle.

Today I got out my geometry tools and with the help of the trapezoid set on the diameter of circle can see that many-- an infinity of irregular hexagons all diagonals intersecting at the circle center and the irregular hexagon center.

Errors listed of 1974 the 7-Circle theorem.
1) error of statement, a IF Then statement with the Then being 6 smaller circles determine 7th larger circle.
2) the theorem produces a 7th circle, it does not produce a intersection point of chords
3) Logically, you can start out with all manner of sizes of 6 smaller circles and only when you demand the constraints of intersection of neighbors and the constraint that chords be diagonals of a irregular hexagon that all meet at one point, can you come to the conclusion that the 6 circles formed a 7th circle.
4) the error of thinking you can start with a large circle, build up 6 smaller circles inside with the contraint of neighbor intersection, and then conclude that all three diagonals of hexagon meet at the center of large circle is untenable.

If nothing else, the AP 7-Circle theorem demands and forces the diagonals of the irregular hexagon meets in the center of the 7th circle. And this is what Wikipedia's page on 7-Circle theorem almost shows, almost shows the intersection is the center of the 7th circle, when in fact, it is the center.

On Thursday, September 1, 2022 at 3:21:11 PM UTC-5, Archimedes Plutonium wrote:
> Alright, I have a proof of the 7-Circle-theorem but it is a theorem totally different from the bullshit theorem which is fake announced by Evelyn, Money-Coutts, and Tyrrell of 1974. Their so-called theorem is in tatters, ruins.
>
> The true theorem associated with 7-Circle involves 6 circles and if the conditions are right, then those 6 circles produces or determines a unique 7th circle that encloses the 6. The 1974 bullshit 7-Circle is focused on an irrelevancy of whether chords intersect at one single point. Whether they do or not is irrelevant to producing a 7th new circle unique to the 6 inner smaller circles.
>
> Say you start out with 6 circles all of different sizes. But not too far different in sizes. For example radius 1, 1.1, 1.2, 1.3, 1.4, 1.5.
>
> Then because they are suitable sizes then form a "somewhat circular chain" etching out a large circle. And suppose their size links all 6 of these circles together, of left and right neighbor intersecting at a single point. If those conditions are met, then we find the most exterior point of each of those 6 circles and connect them up into a hexagon. And that hexagon determines a unique large circle, a 7th circle that encloses all 6 smaller circles.
>
> No-one cares about chords intersection, for the theorem is all concerned about when are 6 smaller circles able to Determine a unique 7th larger circle.
>
> Now I was looking to see if long thin parallelograms form a unique circle encompassing all 4 points. And today I take up that project. Apparently thin long parallelograms are the deciding determinant of whether a IRREGULAR 4-gon determines a unique circle.
>
> All squares and rectangles determine a unique circle. But, it is the parallelogram that at some angle of its **needle like ends, wedge like ends** determines if the enclosed rounded figure will be a circle or the next level of roundness, the ellipse or the next level of roundness the oval.
>
> Here is a picture diagram
>
> /|
> | |
> | |
> |/
>
> You see those needle or wedge ends, if the wedge ends are flat they be a rectangle and allow for a circle creation. But if pointed, they still have symmetry. Can they form a circle? Maybe so, maybe not. But the parallelogram is the deciding figure.
>
> Now if we examine a Trapezoid. And place the base on the diameter of a circle, can all trapezoids determine a circle in this fashion? Some can, some cannot, depending on whether the upper base is reasonably close in length to the lower base.
>
> Now we double the Trapezoid and it forms 6 points of intersection with enclosed circle.
>
> Now we ask the question of diagonal intersection. For these are all Irregular (except one) hexagons. These are similar to the 7-Circle theorem of AP.. And all the diagonals of all these Trapezoids forming irregular hexagons, all of them meet at the circle center as well as the hexagon center.
>
> Now, what if I make these trapezoids, slanted sides different as in the Old Math 7-Circle theorem. And I mirror reflect the trapezoid in the other semicircle. The question here is, does the 3 diagonals meet at one point intersection?
>
> AP

On Thursday, September 1, 2022 at 4:15:15 PM UTC-5, Archimedes Plutonium wrote:
> The AP 7-Circle-Theorem, a correcting of the error filled Old Math 7-Circle-Theorem of 1974.
>
> A statement of the AP 7-Circle-Theorem: The true theorem associated with 7-Circle involves 6 circles and if the conditions are right, then those 6 circles produces or determines a unique 7th circle that encloses the 6. The 1974 bullshit 7-Circle is not focused on the unique determination of the 7th larger circle, and using the intersection as a constraint on Irregular hexagon forming the 7th circle.
>

PROOF of the AP-7-Circle-Theorem:

Proof is easy simple and fast. And students and teachers type in, in a google search that of "parallelogram enclosed in circle"

The first picture they see, or at least I see is a trapezoid inside a semicircle from Math Stack Exchange.

Something like this:

/___\ inside a semicircle with base being the 7th circle.

In my proof it is easy to see I can lower that trapezoid or raise it higher and affecting the three line segment.

I duplicate the trapezoid in other semicircle below, making a full 7th circle.

Thus I form a Irregular hexagon, an infinity of Irregular hexagons that is enclosed by the 7th circle.

Now I fill in each of the 6 sides with a appropriate sized smaller circle, as in the 7-Circle of its 6 smaller circles, with the added stipulation that they must intersect neighbors to right and left.

Now, the question of diagonals of all these infinite irregular hexagons. They all meet in a one point intersection at center of hexagon which is the center of 7th circle.

QED

---------

Proof comments: The Old Math 7-Circle theorem is invalid, it is mired in flaws. It is vague on the intersection of chords. In AP's theorem of 7-Circle, it is crystal clear, that the objective of 7-Circle theorem is to ask the question-- When is a Irregular Hexagon able to form a circumscribing circle? That is the gist, the heart of 7-Circle. The 6 circles are sort of a magicians trick to draw away attention. The 6 circles are what I call bullshit distraction. For the issue of 7Circle theorem is to ask when is IRREGULAR HEXAGON still possessing enough symmetry that it can form and determine a Circle that encloses the irregular hexagon.

AP, King of Science, especially physics

AP seldom reads a proof of mathematics of a famous theorem for that allows AP to prove it himself and go back and see if I or the authors of the proof have it correct. Only when I cannot provide a proof do I read what others have offered.

In the case of 7-Circle-theorem, I went even further. I did not even bother to read the Statement of the Theorem, but conjured up the statement on my own, at my peril. And Wikipedia did not provide either a statement of 7-Circle theorem nor did Wikipedia provide a proof thereof.

Finally I had to go to the world wide web to find out the statement and find out the proof. I still have not read this web site proof. For I know I found the correct proof and that Mr. Rabinowitz's attempt is an illusion. His failure is that the concept and proof of 7-Circle-theorem is all based on hexagons. I doubt he ever mentions hexagons. I do not know how he solves the fact that the intersection of the three chords is actually the center of the large circle.

AP is proud of the idea, that the best way to do mathematics, is do the proofs yourself and not become polluted with someone elses half baked notions and loose logical marbles floating around, tangent or not tangent. (pun intended for the humor impaired reader)

Google search of Seven Circle theorem proof.
--- quoting Google ---
About 20,900,000 results (0.59 seconds) 

Search Results

The Seven Circles Theorem by Stanley Rabinowitz
http://stanleyrabinowitz.com › download › sevenc...
http://stanleyrabinowitz.com › download › sevenc...PDF

by S Rabinowitz · Cited by 5 — The purpose of this article is to give a simple proof of this theorem using only elementary geometry. Since we wish to prove that three lines concur (meet in a ...
10 pages

The Seven Circles Theorem
by Stanley Rabinowitz
Alliant Computer Systems Corporation Littleton, MA 01460
Start with a circle. Any circle. Draw six more circles inside it, each internally tangent to the original circle and tangent to each other in pairs. Let A, B, C, D, E, and F be the consecutive points of tangency of the small circles with the outer circle. We wind up with a set of seven circles as shown in Figure 1. The Seven Circles Theorem says that no matter what sizes we pick for the seven circles (subject only to certain order and tangency constraints), it will turn out that the lines AD, BE, and CF will meet in a point.

--- end quoting Google search ---

Alright, I came to the 7-Circle theorem last month not so much for the mathematics but for the physics of helping to understand the proton torus in 3rd dimension. When I came to the math I supposed/assumed it was correct and valid. But quickly saw that it had errors, and so I devoted almost all the time on fixing the 7-Circle theorem. I am happy to annouce it is fixed and now can get back to the more important part-- physics. The 7-Circle theorem is of course in 2nd dimension, in a plane. But we must not think bad of the 2nd dimension for most of electricity and magnetism is carried out in 2nd dimension, on the surface area.

So as I noted earlier, that 7-Circle theorem was badly misnamed and should be named the Irregular Hexagon theorem for it is about 6 vertices of a hexagon that yield a circle.

And, as I noted earlier on, that the proton torus composed of 8 rings, band rings and the muon 1 band ring. That we can fit in total 4 band rings of muons inside a 8 ring proton torus. In total, we have 12 band rings.

But in physics and in chemistry, that would be 4 electrons inside 1 proton. And that does not make much sense.

So, does the 7-Circle/ Irregular Hexagon theorem have any connection to physics proton torus?

Maybe, for maybe there is a connection. For we know in chemistry there is a Lewis 6 structure, not a Lewis 8 structure. We know that from the strongest bonds in chemistry are CO and not 02. If Lewis 8 structure were true, then O2 would be the strongest bond, not CO, nor N2.

So, here is an idea that comes directly out of the 7-Circle/Irregular Hexagon theorem. In the 8 ring proton torus, each of the 8 rings is a muon itself, and what if 2 of the 8 were to flip from being part of a proton to being a full muon electron. So in hydrogen atom you have a 8 ring proton torus with 1 muon inside going around perpetually doing the Faraday law.

What if 2 of those 8 proton torus rings flipped into being a muon, then we have remaining 6 rings for a proton and 3 rings as muon electrons inside the 6 ring proton.

Here the question arises, could this be the covalent bond of chemistry of the Lewis 6 structure?

Could covalent bonding of chemistry be where proton toruses of 8 rings become toruses of 6 rings and two of the rings convert to muon electrons, inside the 6 ring proton we now have 3 electron muon rings doing the Faraday law..

I do not know and need some time to think about this. To think whether covalent bonding makes better sense as proton rings converting into being muon electrons in the Faraday law.

Of course the Old Chemistry explanation never goes beyond the silly idea of "electron sharing" of the 0.5MeV particles which are actually Dirac magnetic monopoles.

So here, we have a new option for Covalent bonding in chemistry. We have the option of AP's that 2 rings of the 8 ring proton torus become muons doing the Faraday law. Or we have the Old Chemistry option of the Dirac Magnetic Monopoles do the bonding of covalent. There is the problem of the Dirac magnetic monopoles being covalent bonding particles, in that electricity flow is the flow of magnetic monopoles. So the puzzling question occurs as to when the magnetic monopoles flow as current and when they so to speak freeze up and be covalent bonding.

I have to think. But the lessons are quite clear in physics and science-- as we work on one item-- 7-Circles, this helps our minds in uncovering and discovering new ideas in other areas of science.

AP

No, it is not covalent bonding, for that is all under the auspices of the magnetic monopoles. Most of chemistry is under the 0.5MeV particle the Dirac magnetic monopole.

As always I seem to think best in early morning for me, after a night of long sleep.

What the Torus geometry revealed from the 7-Circle theorem is telling us, is that we can peel away 2 of the 8 bands of the proton torus and place those 2 bands inside a 6 band proton torus with 3 bands inside doing the Faraday law.

So, from 8 bands of proton torus with 1 muon inside is normal.

But, we can take 2 of the 8 and place inside, making a 6 band torus with 3 muons inside.

This is not the ordinary usual chemistry. This is the nuclear chemistry and is known as Bremstrahlung radiation. Or is known as Alpha decay as the 2 extra muons inside the 6 band torus come out in Radioactive decay.

I am talking of the heavy elements in the periodic table that engage in switching a 8 band proton torus to a 6 band proton torus.

The radioactive elements that glow in the dark, those are elements where some of its protons have switched from 8 band torus to 6 band torus, and because it has maximum muons-- 3 inside of 6 band proton, we see the Faraday law as this green or blue glow in the dark.

Now I had to get out a huge book, one I bought in early 1990s because of just one picture in that book-- the circular Periodic Table of Elements. The book is Van Nostrand's Scientific Encyclopedia 7th edition Considine, 1989. And on page 2158 is a circular Periodic Table of Chemical Elements.

What the 7-Circle theorem teaches me, is that a torus made of 8 bands can add 4 more bands to the inside of the torus for the Faraday Law. A total of 12 bands for maximum. So for 8 we add 4.

Notice in the Periodic Table that the table starts with 2, the S orbital, then add 4 we have the P orbital of 6. Then add 4 we have the D orbital of 10, then add 4 we have the F orbital of 14, then add 4 we have the G orbital of 18, then add 4 we have the H orbital of 22, then add 4 we have the "i" orbital of 26.

Now how many orbitals is that? That is S,P,D,F,G,H,i for a total of 7 orbitals. Notice that 7-Circles has 7 circles.

Now notice something else, the most stable atom is iron to fusion and fission, and iron is element 26, as in the 7th orbital is "i" at 26.

Notice also that plutonium is the Atom Totality, it has 22 total subshells in 7 shells for 22/7 while only 19 are occupied at any given moment 19/7 is the equiangular number 2.71.... while 22/7 represents pi number 3.14....

Notice also we have H orbital is 22.

But the reason I got out that book is that I can see for myself the Periodic Table in circular form is a torus. And that it is, for it has that donut hole and the elements circling around the donut hole in a logarithmic spiral-- the 2.71.... equiangular spiral.

This is hugely important for all of physics and chemistry, for it begins to tell us how proton band rings are geometrically placed in Space to form a element like plutonium.

The most simple placement for plutonium bands is 94x8 = 752 bands of 105MeV each in one large torus with 94 muon bands inside this 752 proton band. Just one huge torus with 94 electrons inside doing the Faraday law. But that is far too simple of a picture. It is the most simple but not the actual true plutonium atom geometry. I am afraid it gets far far more complicated of a plutonium atom in space. So complex, it looks more like that picture of the circular Periodic Table in Van Nostrand.

AP, King of Science, especially Physics

On Sunday, September 4, 2022 at 11:47:39 AM UTC-5, Ross A. Finlayson wrote:
> If AP's gist is "heart of the 7-circle", I am thinking these are concentric circles, ....
>
> I.e., if it's more "heart of the 7-concentric-circles",
> not "heat of the circle and the six circles centered on it"
>
> What is this 7-circle theorem, I have not really heard of it?
>
> I imagine it might help if some conformal mapping, ...,
> describes flowering or drawing petals in circles.
>
> In reverse, ....

Yes, Ross brings in the topic of whether the 6 interior circles can all be 6 different sizes. And to explore that question, let me draw up my full proof of September 1, 2022, which talks about the heart or gist of the proof. For the constraint that all three chords meet at one point, forces the circles to __not all 6 different sizes___. This AP proof is a fine fine example of where Computer Graphics ruins and destroys the truth of geometry. Also that destruction is seen in the silly daft Goldberg polyhedrons where Goldberg has 7 hexagons, 6hexagons surrounding a central 7 hexagons tiling the surface of sphere. Had Goldberg gone crazy???? For 7hexagons with a central hexagon is still a PLANE. In some of Goldberg's writings he is talking about "hinges, and ball and socket joints", again, had Goldberg gone crazy, yet that nonsense was published.

And the ruination of Geometry math can be seen with Internet pictures-- Wikipedia of Goldberg polyhedron with 7 hexagons, 7th central to 6 surrounding hexagons.

The ruination of geometry proving is seen in the silly prattle of Appell & Haken 4 Color Mapping and the disturbing idiocy of Thomas Hales computer fake proof of Kepler Packing.

It is fair to say-- computer graphics can destroy any geometry proof with a computer graphic that is the total opposite of what the proof statement was.

One can see it in the Dandeloin silly fakery where a ellipse is inserted inside a cone and two spheres are juggled. Impossible in Reality, for the oval is the slant cut of cone, but in Computer Graphics of modern day-- any true statement of geometry can be totally destroyed by computer graphics.

On Thursday, September 1, 2022 at 4:15:15 PM UTC-5, Archimedes Plutonium wrote:
> The AP 7-Circle-Theorem, a correcting of the error filled Old Math 7-Circle-Theorem of 1974.
>
> A statement of the AP 7-Circle-Theorem: The true theorem associated with 7-Circle involves 6 circles and if the conditions are right, then those 6 circles produces or determines a unique 7th circle that encloses the 6. The 1974 bullshit 7-Circle is not focused on the unique determination of the 7th larger circle, and using the intersection as a constraint on Irregular hexagon forming the 7th circle.
>

PROOF of the AP-7-Circle-Theorem:

Proof is easy simple and fast. And students and teachers type in, in a google search that of "parallelogram enclosed in circle"

The first picture they see, or at least I see is a trapezoid inside a semicircle from Math Stack Exchange.

Something like this:

/___\ inside a semicircle with base being the 7th circle.

In my proof it is easy to see I can lower that trapezoid or raise it higher and affecting the three line segment.

I duplicate the trapezoid in other semicircle below, making a full 7th circle.

Thus I form a Irregular hexagon, an infinity of Irregular hexagons that is enclosed by the 7th circle.

Now I fill in each of the 6 sides with a appropriate sized smaller circle, as in the 7-Circle of its 6 smaller circles, with the added stipulation that they must intersect neighbors to right and left.

Now, the question of diagonals of all these infinite irregular hexagons. They all meet in a one point intersection at center of hexagon which is the center of 7th circle.

QED

---------

Proof comments: The Old Math 7-Circle theorem is invalid, it is mired in flaws. It is vague on the intersection of chords. In AP's theorem of 7-Circle, it is crystal clear, that the objective of 7-Circle theorem is to ask the question-- When is a Irregular Hexagon able to form a circumscribing circle? That is the gist, the heart of 7-Circle. The 6 circles are sort of a magicians trick to draw away attention. The 6 circles are what I call bullshit distraction. For the issue of 7Circle theorem is to ask when is IRREGULAR HEXAGON still possessing enough symmetry that it can form and determine a Circle that encloses the irregular hexagon.

On 4Sept2022, AP writes:

It is quick and easy, and you need not be exacting or precise. So get a round object and with pencil draw it on a scrap paper. Draw a diameter, then draw a trapezoid in northern hemisphere

/___\

Now draw the same trapezoid in southern hemisphere, turned around.

Now connect all 3 diagonals and they meet at the center of the circle.

Now suppose we wanted 6 different sized circles for the interior 6 circles. Are we able to use a different trapezoid for southern hemisphere than the northern hemisphere trapezoid.

So, make the southern hemisphere trapezoid be different from northern. And then look to see of the 3 diagonals? Here, we find out that if the two trapezoids are _not identical_ then the 3 diagonals do not meet in a single point intersection.

This suggests, that the size of the 6 interior circles can only vary in 3 different sizes, not in 6 different sizes. That 3 circles opposite the other 3 must be equal in size in order for the diagonals to meet at the center of the 7th circle.

AP

The 6 interior circles have to meet at a single point intersection, otherwise they violate the condition of tangent to left and right neighbor. The only point of intersection for the 3 chords of the 6 circles can only be the 7th circle center. You cannot have 3 point tangency of two large circles and a small tiny one wedged between. The neighbor tangency can only be right and left neighbor, not threesome tangent.

If we relax the condition that the northern hemisphere trapezoid be different from southern hemisphere trapezoid, we only cause the diagonals to not meet at a one point intersection. And there is no point in exploring other options of "more anti-symmetry" keeping in mind, this is a circle, and requiring absolute symmetry. So if we cannot make two different hemisphere trapezoids obey 7-Circle. We sure as hell are not going to make antisymmetry more loss of symmetry in a quadrilateral of random points in that hemisphere obey 7-Circle.

The constraints are clear--- Left to Right tangency only, and a intersection of the diagonals (chords) as the center of the 7th circle. This is what makes the 1974 by Evelyn, Money-Coutts, and Tyrrell an invalid statement of geometry and the proofs such as Rabinowitz offering as invalid-- that the intersection has to be the 7th circle center.

The 7-Circle theorem is more a theorem about when a Irregular Hexagon can be enclosed by 1 large circle than it ever is a theorem about 6 smaller interior circles.

And the statement of the theorem should reflect that idea.

Here is what a more logical statement of 7-Circle would say:

Given a trapezoid whose base is the diameter of a circle, then all irregular hexagons that can be circumscribed from that diameter are the infinite number of two identical trapezoids formed into irregular hexagon as shown in picture. There is one set of trapezoids that forms a Regular Hexagon, and the infinite others are Irregular hexagons.

/____\
\____/

The error of 1974 by Evelyn, Money-Coutts, and Tyrrell, is they are vague, cloudy, fuzzy on the intersection of the 3 chords. So vague they thought you can have 6 different sizes for the interior circles. Which is not true. You can have 3 different sizes.

But 1974 is a ripe time for Computer Graphics Art to enter geometry, and where the computer makes a Optical Illusions of everything in geometry. The wikipedia page in 7-Circle theorem is a grotesque optical illusion of geometry.

AP, King of Science, especially Physics

Mathematics needed a theorem that tells when a Irregular Hexagon makes a circumscribing circle of its vertices. Unfortunately, that theorem came with silly baggage of interior circles. The answer to when does a Irregular Hexagon allow to be insribed inside a big circle, is when the hexagon is two identical trapezoids with their base in common.

As to filling up the big circle with smaller circles is petty information, and is a lemma of the overarching theorem that two identical trapezoids on the same base forms a large circle for its 6 vertices. Of course the diagonals intersect in the circles center. But Old Math had the 7-Circle theorem a clusterfuck theorem, vague on the intersection, vague on circle sizes, and forgot to even mention how crucial the trapezoids are.

And Old Math's 7-Circle theorem comes at a time in math history that computer graphics can deliver a picture of any geometry figure that is 180degrees opposite of the truth. In other words, computer graphics have become a hindrance to the truth of geometry-- just look at the silly Goldberg polyhedron in Wikipedia or Google Search where 6 hexagons surround a middle hexagon and tile the sphere surface. Goldberg published in a math journal-- a gutter press journal that tiling hexagons can have a swivel hinge and ball and socket joints. Who knows what Goldberg and the math publishers were smoking that day.

AP

Totally fascinating that there seems to be no Octagon circle to compare to the Hexagon circle of 7-Circle-theorem.

The Trapezoid-- is a very very interesting and important figure of mathematics geometry. For it is like a isosceles triangle with its upper part cut and removed with two opposite sides parallel and the side lengths of same angle.

So, playing with this in your mind, for I have not played with this in my mind when doing High School Geometry.

So a trapezoid is like this

/____\

Now we flip that second side at the angle of first to get this
di
/_____/

And now we have instead a parallelogram.

So I was not consciously aware of that feature of trapezoids and how special they are.

So now I go a step further and wonder about a 9-Circle theorem involving octagons inside a 9th large circle, instead of Irregular Hexagons in 7-Circle theorem.

But here I am stuck, for I cannot think of the next step of a trapezoid to make for octagons in the 9-Circle Theorem.

Would it be Pentagons? Where one side is the 9th circle diameter, that would leave 4 sides in Northern Hemisphere. And when I place the identical pentagon in Southern hemisphere, that should form a Octagon.

Trouble is, though the shared base has to be much larger than the other four sides of the pentagon to be the diameter of the 9th largest circle.

Now the diagonals would be 4 diagonals and if the 4 sides are all equal, then they pass through the center of the 9th circle. However, I cannot imagine how I can get different size circles inside the 9th circle.

So is the case of the 9th circle theorem a study of one unique case? Apparently it is.

AP

I need to start over completely here.

For this circle arranging inside a large circle and a chain of interior smaller circle is important in analysis of division evenly of pi. This logically leads to the infinity borderline at 1.2*10^604.

Now the prefix digits of 1.2 are important for physics constants, and in particular for Magnetic Permeability 1.2*10^-6 along with Electric Permittivity of 8.4*10^-12.

The prefix of 1.2 comes from 5 factorial = 120. But let us see how it plays out in geometry and for seeking out the infinity borderline in geometry by chains of smaller circles inside a large circle as in 7-Circle theorem of Old Math, and AP's correction of that tarnished theorem in New Math.

Previously I said this borderline was the number 1*10^604.

Well, as time passes we learn more and more, for that number is not as precise as 1.2*10^604.

Pi is evenly divisible by 2 for you can make semicircles; by 3 for the equilateral triangle is circumscribed; by 4 and 5 for the same reason of circumscribed regular polygons. Pi is evenly divisible by 5 factorial = 2*3*4*5 = 120.

Since the Circle is evenly divisible by 2, it is evenly divisible by 2, 4, 8, 16, 32, ....
Since it is divisible by 3, hence divisible by 3, 6, 12, 24, 48,..... and by 3, 9, 27, .....
Since it is evenly divisible by 5, it is also evenly divisible by 5, 10, 20, 40, .....

AP